If M is a finite monoid and in M implies that , show that is a unit.
with hints: If show that are distinct.
If then by definition it has an identity element . Then, because M is finite, there will be an element which will map to the identity of M. Each will be distinct as for every , u will map it back to itself. Thus, which, because will map to itself and so must be a unit.
Oct 22nd 2009, 04:50 AM
What you have to prove is:
To show the hint, just use that (for instance prove it by contraposition).
Assume you've proved it. Show that (using a property of finite sets)
Therefore means . Can you conclude?
Oct 22nd 2009, 10:10 PM
would it suffice to create a cayley table with elements 1,a,b and show that a and b both have an inverse? ugh, where would i go from there?
thanks, really appreciate it
Oct 23rd 2009, 12:12 AM
I don't really understand, why just in a Cayley table. can be different from or
What the proof says is : the condition you have implies that the map is injective and then surjective (because it's an injective map between sets of same finite cardinality) therefore has an antecedant, which is some such that and we're done.